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The free body diagram in shown below:
From the diagram and the problem we have that:
• The weight and angle of the inclined plane are given.
,• The normal force and the components of the weight are unknown (this implies that the acceleration is unknown too); we also notice that the mass is not given the it is also an unknown.
We know that Newton's second law states that:
[tex]\vec{F}=m\vec{a}[/tex]where F is the resultant force and a is the acceleration. Since this is a vector equation we can decomposed it in two scalar equations (in this case we only need two scalar equations since the forces are coplanar), then we have:
[tex]\begin{gathered} Wx=ma_x \\ N-W_y=ma_y \end{gathered}[/tex]Since we don't expect the cyclist to move in the y direction (otherwise he will surely fall) the equations above would reduce to:
[tex]\begin{gathered} W_x=ma \\ N-W_y=0 \end{gathered}[/tex]From the first equation we can solve the acceleration, to do this we use the triangle to get the x-component of the weight:
[tex]\begin{gathered} W_x=ma \\ W\sin \theta=ma \\ a=\frac{W\sin \theta}{m} \end{gathered}[/tex]Since the weight is given but not the mass we use the fact that the weight is:
[tex]W=mg[/tex]to get the mass, then we have:
[tex]\begin{gathered} m=\frac{W}{g} \\ m=\frac{75}{9.8} \\ m=7.65 \end{gathered}[/tex]hence the mass of the cylcist is 7.65 kg.
Now that we have all the values we need we plug them in the expression for the acceleration:
[tex]\begin{gathered} a=\frac{75\sin 7}{7.65} \\ a=1.19 \end{gathered}[/tex]Therefore the acceleration of the cyclist is 1.19 meters per second per second.
